# Time Reversal in Euclidean Spacetime - unitary or antiunitary?

+ 4 like - 0 dislike
296 views

(pre-request) We know that time reversal operator $T$ is an anti-unitary operator in Minkowsi Spacetime. i.e.

$$T z=z^*T$$ where the complex number $z$ becomes its complex conjugate. See, for example, Peskin and Schroeder An Introduction To Quantum Field Theory'' p.67 Eq (3.133).

(Question) Is Euclidean time reversal operator $T_E$ an unitary operator, i.e.

$$T_E z=z T_E \;\;\;(?)$$

or an anti-unitary operator in Euclidean Spacetime? Why is necessarily that or why is it necessarily not that? Or should $T_E$ be an unitary operator like Parity $P$ instead?

However, here see the attempt of a PRL paper, Euclidean continuation of the Dirac fermion where time reversal $T_E$ is given by, on page 3: $$T_E z=z^*T_E \;\;\;(?)$$ $T_E$ is still an anti-unitary operator!

It seems to me if one consider Euclidean spacetime, the Euclidean signature is the same sign, say $(-,-,-,-,\dots)$, then Parity $P$ acts on the space is equivalent to the Euclidean time reversal operator $T_E$ acts on Euclidean time (which is now like one of the spatial dimensions). So shouldn't $T_E$ be an unitary operator as Parity $P$?

[Other Refs]

i. Please, you may read, an earlier Phys.SE question here, probably is poorly formulated, so cannot draw the efforts of people to answer the question. Here let me try an easier way and focus on one issue only.

ii. Osterwalder-Schrader (OS) approach, and A continuous Wick rotation for spinor fields and supersymmetry in Euclidean space'' by Peter van Nieuwenhuizen, Andrew Waldron.

This post imported from StackExchange Physics at 2014-06-04 11:36 (UCT), posted by SE-user Idear
Starting with $Tz = z^*T$, and assuming $T = iT_E$, this would give $(iT_E)z = z^*(iT_E)$, that is $i(T_Ez) = (z^*i)T_E$, that is $(T_Ez) = -i(z^*i)T_E$, that is $T_Ez =z^*T_E$
Thanks Trimok, this can be an mathematical answer for a logical reason; I am sure that is what have been done in wick rotating with fermion fields. So even if Euclidean spacetime seemly treat spatial and euclidean-time the same in the signature/metric sense, but when the fermion gets involved, it will still tell the difference between euclidean-time'' and the remained space''? Simply by this anti-unitary property?
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.